When any number is divided by 5, the remainder can only be 0, 1, 2, 3 or 4. The remainder cannot exceed the divisor. So perhaps the question is worded wrongly.
When any number is divided by a number exceeding 20 then the remainder goes from 0 to one less than the number itself. So perhaps the question should have been: find 3 numbers between 1000 and 1200 that have a remainder of 20 when divided by a multiple of 5.
The multiple of 5 can be called 5n where n>4. So each number, x, has to satisfy the equation:
x/(5n)=y+20/(5n); so x=5ny+20, where 1000≤5ny+20≤1200.
That is, 980≤5ny≤1180; 196≤ny≤236. Both n and y must be integers, and n>4.
But there's an upper limit to n, because we have to have at least 3 integer values to choose from for y. This limit is n=14, because if we put n=14, 14≤y≤16, so y=14, 15 or 16. Therefore 4<n≤14.
Let's use n=13 and see what we have for y: 15<y<18, i.e., y=16, 17 or 18. When n=13 the divisor is 5*13=65 and the numbers are 5ny+20=65*16+20=1060; 65*17+20=1125; 65*18+20=1190.
Now we need to check that the numbers 1060, 1125 and 1190 divided by 65 give us the expected results.
1060/65=16 rem 20; 1125/65=17 rem 20; 1190/65=18 rem 20.
There are many more possible solutions, assuming the correction to the question is correct.
One other type of solution produces the numbers 1000, 1100, 1200: 5*196+20=1000; 5*216+20=1100; 5*236+20=1200. In this example solution, the chosen numbers when divided by another number give 5 as the quotient with remainder 20. This is another interpretation of the question. 1005, 1100 and 1195 also work: 5*197+20=1005; 5*216+20=1100; 5*235+20=1195.